Transcript The Foundations: Logic and Proofs

Chapter 1, Part III: Proofs
With Question/Answer Animations
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Summary
 Valid Arguments and Rules of Inference
 Proof Methods
 Proof Strategies
Section 1.6
Section Summary
 Valid Arguments
 Inference Rules for Propositional Logic
 Using Rules of Inference to Build Arguments
 Rules of Inference for Quantified Statements
 Building Arguments for Quantified Statements
Revisiting the Socrates Example
 We have the two premises:
 “All men are mortal.”
 “Socrates is a man.”
 And the conclusion:
 “Socrates is mortal.”
 How do we get the conclusion from the premises?
The Argument
 We can express the premises (above the line) and the
conclusion (below the line) in predicate logic as an
argument:
 We will see shortly that this is a valid argument.
Valid Arguments

We will show how to construct valid arguments in
two stages; first for propositional logic and then for
predicate logic. The rules of inference are the
essential building block in the construction of valid
arguments.
1.
Propositional Logic
Inference Rules
2.
Predicate Logic
Inference rules for propositional logic plus additional inference
rules to handle variables and quantifiers.
Arguments in Propositional Logic
 A argument in propositional logic is a sequence of propositions.
All but the final proposition are called premises. The last
statement is the conclusion.
 The argument is valid if the premises imply the conclusion. An
argument form is an argument that is valid no matter what
propositions are substituted into its propositional variables.
 If the premises are p1 ,p2, …,pn and the conclusion is q then
(p1 ∧ p2 ∧ … ∧ pn ) → q is a tautology.
 Inference rules are all argument simple argument forms that will
be used to construct more complex argument forms.
Rules of Inference for Propositional
Logic: Modus Ponens
Corresponding Tautology:
(p ∧ (p →q)) → q
Example:
Let p be “It is snowing.”
Let q be “I will study discrete math.”
“If it is snowing, then I will study discrete math.”
“It is snowing.”
“Therefore , I will study discrete math.”
Modus Tollens
Corresponding Tautology:
(¬q∧(p →q))→¬p
Example:
Let p be “it is snowing.”
Let q be “I will study discrete math.”
“If it is snowing, then I will study discrete math.”
“I will not study discrete math.”
“Therefore , it is not snowing.”
Hypothetical Syllogism
Corresponding Tautology:
((p →q) ∧ (q→r))→(p→ r)
Example:
Let p be “it snows.”
Let q be “I will study discrete math.”
Let r be “I will get an A.”
“If it snows, then I will study discrete math.”
“If I study discrete math, I will get an A.”
“Therefore , If it snows, I will get an A.”
Disjunctive Syllogism
Corresponding Tautology:
(¬p∧(p ∨q))→q
Example:
Let p be “I will study discrete math.”
Let q be “I will study English literature.”
“I will study discrete math or I will study English literature.”
“I will not study discrete math.”
“Therefore , I will study English literature.”
Addition
Corresponding Tautology:
p →(p ∨q)
Example:
Let p be “I will study discrete math.”
Let q be “I will visit Las Vegas.”
“I will study discrete math.”
“Therefore, I will study discrete math or I will visit
Las Vegas.”
Simplification
Corresponding Tautology:
(p∧q) →p
Example:
Let p be “I will study discrete math.”
Let q be “I will study English literature.”
“I will study discrete math and English literature”
“Therefore, I will study discrete math.”
Conjunction
Corresponding Tautology:
((p) ∧ (q)) →(p ∧ q)
Example:
Let p be “I will study discrete math.”
Let q be “I will study English literature.”
“I will study discrete math.”
“I will study English literature.”
“Therefore, I will study discrete math and I will study
English literature.”
Resolution
Resolution plays an important role
in AI and is used in Prolog.
Corresponding Tautology:
((¬p ∨ r ) ∧ (p ∨ q)) →(q ∨ r)
Example:
Let p be “I will study discrete math.”
Let r be “I will study English literature.”
Let q be “I will study databases.”
“I will not study discrete math or I will study English literature.”
“I will study discrete math or I will study databases.”
“Therefore, I will study databases or I will study English
literature.”
Using the Rules of Inference to
Build Valid Arguments
 A valid argument is a sequence of statements. Each statement is
either a premise or follows from previous statements by rules of
inference. The last statement is called conclusion.
 A valid argument takes the following form:
S1
S2
.
.
.
Sn
C
Valid Arguments
Example 1: From the single proposition
Show that q is a conclusion.
Solution:
Valid Arguments
Example 2:
 With these hypotheses:
“It is not sunny this afternoon and it is colder than yesterday.”
“We will go swimming only if it is sunny.”
“If we do not go swimming, then we will take a canoe trip.”
“If we take a canoe trip, then we will be home by sunset.”
 Using the inference rules, construct a valid argument for the conclusion:
“We will be home by sunset.”
Solution:
1.
Choose propositional variables:
p : “It is sunny this afternoon.” r : “We will go swimming.” t : “We will be home by sunset.”
q : “It is colder than yesterday.” s : “We will take a canoe trip.”
2. Translation into propositional logic:
Continued on next slide 
Valid Arguments
3. Construct the Valid Argument
Handling Quantified Statements
 Valid arguments for quantified statements are a
sequence of statements. Each statement is either a
premise or follows from previous statements by rules
of inference which include:
 Rules of Inference for Propositional Logic
 Rules of Inference for Quantified Statements
 The rules of inference for quantified statements are
introduced in the next several slides.
Universal Instantiation (UI)
Example:
Our domain consists of all dogs and Fido is a dog.
“All dogs are cuddly.”
“Therefore, Fido is cuddly.”
Universal Generalization (UG)
Used often implicitly in Mathematical Proofs.
Existential Instantiation (EI)
Example:
“There is someone who got an A in the course.”
“Let’s call her a and say that a got an A”
Existential Generalization (EG)
Example:
“Michelle got an A in the class.”
“Therefore, someone got an A in the class.”
Using Rules of Inference
Example 1: Using the rules of inference, construct a valid
argument to show that
“John Smith has two legs”
is a consequence of the premises:
“Every man has two legs.” “John Smith is a man.”
Solution: Let M(x) denote “x is a man” and L(x) “ x has two legs”
and let John Smith be a member of the domain.
Valid Argument:
Using Rules of Inference
Example 2: Use the rules of inference to construct a valid argument
showing that the conclusion
“Someone who passed the first exam has not read the book.”
follows from the premises
“A student in this class has not read the book.”
“Everyone in this class passed the first exam.”
Solution: Let C(x) denote “x is in this class,” B(x) denote “ x has read
the book,” and P(x) denote “x passed the first exam.”
First we translate the
premises and conclusion
into symbolic form.
Continued on next slide 
Using Rules of Inference
Valid Argument:
Returning to the Socrates Example
Solution for Socrates Example
Valid Argument
Universal Modus Ponens
Universal Modus Ponens combines universal
instantiation and modus ponens into one rule.
This rule could be used in the Socrates example.
Section 1.7
Section Summary
 Mathematical Proofs
 Forms of Theorems
 Direct Proofs
 Indirect Proofs
 Proof of the Contrapositive
 Proof by Contradiction
Proofs of Mathematical Statements
 A proof is a valid argument that establishes the truth of a
statement.
 In math, CS, and other disciplines, informal proofs which are
generally shorter, are generally used.





More than one rule of inference are often used in a step.
Steps may be skipped.
The rules of inference used are not explicitly stated.
Easier for to understand and to explain to people.
But it is also easier to introduce errors.
 Proofs have many practical applications:
 verification that computer programs are correct
 establishing that operating systems are secure
 enabling programs to make inferences in artificial intelligence
 showing that system specifications are consistent
Definitions
 A theorem is a statement that can be shown to be true using:
 definitions
 other theorems
 axioms (statements which are given as true)
 rules of inference
 A lemma is a ‘helping theorem’ or a result which is needed to
prove a theorem.
 A corollary is a result which follows directly from a theorem.
 Less important theorems are sometimes called propositions.
 A conjecture is a statement that is being proposed to be true.
Once a proof of a conjecture is found, it becomes a theorem. It
may turn out to be false.
Forms of Theorems
 Many theorems assert that a property holds for all elements
in a domain, such as the integers, the real numbers, or
some of the discrete structures that we will study in this
class.
 Often the universal quantifier (needed for a precise
statement of a theorem) is omitted by standard
mathematical convention.
For example, the statement:
“If x > y, where x and y are positive real numbers, then x2 > y2 ”
really means
“For all positive real numbers x and y, if x > y, then x2 > y2 .”
Proving Theorems
 Many theorems have the form:
 To prove them, we show that where c is an arbitrary
element of the domain,
 By universal generalization the truth of the original
formula follows.
 So, we must prove something of the form:
Proving Conditional Statements: p → q
 Trivial Proof: If we know q is true, then
p → q is true as well.
“If it is raining then 1=1.”
 Vacuous Proof: If we know p is false then
p → q is true as well.
“If I am both rich and poor then 2 + 2 = 5.”
[ Even though these examples seem silly, both trivial and vacuous
proofs are often used in mathematical induction, as we will see
in Chapter 5) ]
Even and Odd Integers
Definition: The integer n is even if there exists an
integer k such that n = 2k, and n is odd if there exists
an integer k, such that n = 2k + 1. Note that every
integer is either even or odd and no integer is both
even and odd.
We will need this basic fact about the integers in some
of the example proofs to follow. We will learn more
about the integers in Chapter 4.
Proving Conditional Statements: p → q
 Direct Proof: Assume that p is true. Use rules of inference,
axioms, and logical equivalences to show that q must also
be true.
Example: Give a direct proof of the theorem “If n is an odd
integer, then n2 is odd.”
Solution: Assume that n is odd. Then n = 2k + 1 for an
integer k. Squaring both sides of the equation, we get:
n2 = (2k + 1)2 = 4k2 + 4k +1 = 2(2k2 + 2k) + 1= 2r + 1,
where r = 2k2 + 2k , an integer.
We have proved that if n is an odd integer, then n2 is an
odd integer.
( marks the end of the proof. Sometimes QED is
used instead. )
Proving Conditional Statements: p → q
Definition: The real number r is rational if there exist
integers p and q where q≠0 such that r = p/q
Example: Prove that the sum of two rational numbers
is rational.
Solution: Assume r and s are two rational numbers.
Then there must be integers p, q and also t, u such
that
where v = pu + qt
w = qu ≠ 0
Thus the sum is rational.
Proving Conditional Statements: p → q
 Proof by Contraposition: Assume ¬q and show ¬p is true also. This is
sometimes called an indirect proof method. If we give a direct proof of
¬q → ¬p then we have a proof of p → q.
Why does this work?
Example: Prove that if n is an integer and 3n + 2 is odd, then n is
odd.
Solution: Assume n is even. So, n = 2k for some integer k. Thus
3n + 2 = 3(2k) + 2 =6k +2 = 2(3k + 1) = 2j for j = 3k +1
Therefore 3n + 2 is even. Since we have shown ¬q → ¬p , p → q
must hold as well. If n is an integer and 3n + 2 is odd (not even) ,
then n is odd (not even).
Proving Conditional Statements: p → q
Example: Prove that for an integer n, if n2 is odd, then n is
odd.
Solution: Use proof by contraposition. Assume n is even
(i.e., not odd). Therefore, there exists an integer k such
that n = 2k. Hence,
n2 = 4k2 = 2 (2k2)
and n2 is even(i.e., not odd).
We have shown that if n is an even integer, then n2 is even.
Therefore by contraposition, for an integer n, if n2 is odd,
then n is odd.
Proving Conditional Statements: p → q
 Proof by Contradiction: (AKA reductio ad absurdum).
To prove p, assume ¬p and derive a contradiction such as
p ∧ ¬p. (an indirect form of proof). Since we have shown
that ¬p →F is true , it follows that the contrapositive T→p
also holds.
Example: Prove that if you pick 22 days from the calendar,
at least 4 must fall on the same day of the week.
Solution: Assume that no more than 3 of the 22 days fall
on the same day of the week. Because there are 7 days of
the week, we could only have picked 21 days. This
contradicts the assumption that we have picked 22 days.
Proof by Contradiction
 A preview of Chapter 4.
Example: Use a proof by contradiction to give a proof that √2 is
irrational.
Solution: Suppose √2 is rational. Then there exists integers a and b
with √2 = a/b, where b≠ 0 and a and b have no common factors (see
Chapter 4). Then
Therefore a2 must be even. If a2 is even then a must be even (an
exercise). Since a is even, a = 2c for some integer c. Thus,
Therefore b2 is even. Again then b must be even as well.
But then 2 must divide both a and b. This contradicts our assumption
that a and b have no common factors. We have proved by contradiction
that our initial assumption must be false and therefore √2 is
irrational .
Proof by Contradiction
 A preview of Chapter 4.
Example: Prove that there is no largest prime number.
Solution: Assume that there is a largest prime
number. Call it pn. Hence, we can list all the primes
2,3,.., pn. Form
None of the prime numbers on the list divides r.
Therefore, by a theorem in Chapter 4, either r is prime
or there is a smaller prime that divides r. This
contradicts the assumption that there is a largest
prime. Therefore, there is no largest prime.
Theorems that are Biconditional
Statements
 To prove a theorem that is a biconditional statement,
that is, a statement of the form p ↔ q, we show that
p → q and q →p are both true.
Example: Prove the theorem: “If n is an integer, then n is
odd if and only if n2 is odd.”
Solution: We have already shown (previous slides) that
both p →q and q →p. Therefore we can conclude p ↔ q.
Sometimes iff is used as an abbreviation for “if an only if,” as in
“If n is an integer, then n is odd iif n2 is odd.”
What is wrong with this?
“Proof” that 1 = 2
Solution: Step 5. a - b = 0 by the premise and
division by 0 is undefined.